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coprime, coprime there is no.1#include 2#include 3 using namespacestd;4 5 BOOLfunc (__int64 a,__int64 b)6 {7 __int64 temp;8 if(ab)9 {Tentemp=A; OneA=b; Ab=temp; - } -__int64 c=a%b; the while(c!=0) - { -c=a%b; -A=b; +b=C; - } + if(a==1)return 1;//Huzhi A return 0; at } - - intMain () - { - intT; - __int64 m,n; inscanf"%d",t); - while(t--) to { +scanf"%i64d%i64d",m,n); - if(m==1|| n==1) the { *printf"no\n"); $ Continue;Pan

Title Link: http://acm.hdu.edu.cn/showproblem.php?pid=2588Title: The number of all numbers with a title that requires less than or equal to N and greatest common divisor of n is greater than or equal to MMethod: Using Euler functionsSo how do you write it? First the Euler function can only find the number of all the numbers that are less than N and N coprime, but there are the numbers we need, so we can mak

(scanf ("%d", n)! = EOF) {for (int i = 0; i A[i] = 1;scanf ("%d%d", m[i], b[i]);}PLL ans = Linear (A, B, M, n);if (Ans.second = =-1) printf (" -1\n");else printf ("%i64d\n", Ans.first);}} 1 #include #include #include using namespace Std;typedef long Long LL;typedef pairLL a[100000], b[100000], m[100000];ll GCD (ll A, ll b) {Return b? GCD (b, a%b): A;}void Ex_gcd (ll A, ll B, LL x, LL y, ll d) {if (!b) {d = a, x = 1, y = 0;}else{EX_GCD (b, a% B, y, X, D);Y-= x * (A/b);}}LL INV (ll T, ll P) {//If

(A,aa[i],x,y);//Ax-aa[i]y=bb[i]-b * if((bb[i]-b)%d!=0) {ok=true; Break;} $X=x* ((bb[i]-b)/d);Panax Notoginseng -LL T=mabs (aa[i]/d); thex= (x%t+t)%T; + Ab=a*x+b,a=a*aa[i]/d;//LCM (A,aa[i]); the } + if(OK) printf ("0\n"); - Else $ { $LL ans= (b%a+a)%a,cnt=0; //x=ax+b The minimum nonnegative integer solution for the X at the time of ans - if(ans>0 ans; // if ANS is eligible for X-value to be counted in CNT -cnt+= (N-ans)/A; The

.$$ solution: $$ (A, b) = 1\rightarrow (a^2, b) = 1 \rig Htarrow (a^2 + b^2, b) = 1.$$ similarly available $ (a^2 + b^2, a) = 1$. So $ (a^2 + b^2, AB) = 1$.5. Proof: If $a, b\in\mathbf{n^*}$, then sequence: $a $, $2a$, $\cdots$, $ba $ The number of items that can be divisible by $b $ equals $ (A, b) $.Answer:Make $ (A, B) = d$, $a = da_1$, $b = db_1$, $ (a_1, b_1) = 1$. The original number is listed as $ $

) 1e9 +7;4 intF[n], Finv[n], inv[n];//f is factorial, FINV is the factorial of inverse element5 voidinit () {6inv[1] =1;7 for(inti =2; i ){8Inv[i] = (mod-mod/i) * 1LL * inv[mod% i]%MOD;9 }Tenf[0] = finv[0] =1; One for(inti =1; i ){ AF[i] = f[i-1] * 1LL * I%MOD; -Finv[i] = finv[i-1] * 1ll * Inv[i]%MOD; - } the } - intComb (intNintm) {//comb (n, m) is C (n, m) - if(M 0|| M > N)return 0; - returnF[N] * 1ll * finv[n-m]% MOD * Finv[m]%MOD; + } - intMain () { + init (); A}The

(); for(inti =1; I -; i + +) {cout" "; }return 0;}The linear sieve method to find the Möbius functionCode:#include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #define LL Long Long#define INF 0x7fffffffusing namespace STD;Const intMAXN =1000005;BOOLVIS[MAXN];intPRIME[MAXN];intMU[MAXN];//Möbius functionintTotvoidInit () {memset(Vis,false,sizeof(VIS)); mu[1] =1;

The main topic: given a 2^e prefix N, the number of bits of the known prefix n is less than half the number of bits of 2^e, find the minimum e that satisfies the condition.Problem Analysis: Set 2^e as I digit, then have n*10^iThe code is as follows:# include# include# include# include# includeusing namespacestd;voidWork (unsigned n) {DoubleU= (Double) n; intI=LOG10 (U) +2; while(1){ intLow=floor

UVA 1434-yaptchaTopic linksTest instructions: Find out the answer to the formula in the questionIdeas:When 3? k + 7 Non-prime time, then (3? ) k + 6 )! (Two factors must be found to multiply)So the original is 0When 3? k + 7 As a prime number, according to the Wilson theorem,((3?k+6)!+1)%(3?k+7)==0, so the primitive can be converted to [x-(x-1)] = 1So the question turns to just to infer that 3 * k + 7 is not a prime

each pair of adjacent pairs, becoming 0/1,1/4,1/3, 2/5, 1/2, 3/5,2/3, 3/4 , 1/1 Between STEP5:0/1 and 1/4 and 3/4 and 1/1 can still be inserted 1 numbers, and the number of inserted denominator is not greater than 5 0/1, 1/5 , Quarter , 1/3, 2/5, , 3/5, 2/3, 3/4,4/5 , 1/1 At this point, the sequence contains all of the least-true fractions with a denominator of not more than 5, and the individual fractions are arranged in ascending order.1#

DescriptionTo solve the equations of linear equation, \ (m_i\) is not coprime.SolExpand Euclid + Chinese remainder theorem.The first 22 merges with the same blog post.Each general solution is to increase the number of least common multiple each time two, this modulo any number is 0.The pseudo code is as followsM = m[1], R = r[1]for i = 2.. N d = gcd (M, m[i]) c = r[i]-RIf (c mod d) then//without solution re

make "[(the initial value of N) + 1] % K "is equal to" (the final value of N) % K ". the second linePrint the operations to do in each step, which consist of '+', '-', '*' and '% '. if there are more than one solution, print the minimum one. (Here we define '+' And only if there exists a p such that for I = 1,..., P-1, AI = Bi, and for I = P, AI Sample Input 2 2 2-1 12 100 0 0 Sample output 02*+Meaning: (Note that % in the question refers to mod) I gave you n, k, m .... Each time a new N is ob

2014 Multi-University Training Contest 1/HDU4861_Couple doubi (number theory/law), hdu4861The problem-solving report shows that the two take the ball in turn, and the big ones win, and post an official question, and, anyway, I don't understand, and, first, keep understanding about the ferma small theorem about the original root.Find, sad, and the cycle section P-1. Each cycle contains a non-zero ball. There

Topic: Given n, to find LCM (1,n) +LCM (2,n) +...+LCM (n,n)Enumeration D=GCD (I,n), so that f (n) is n and the sum of n coprimeThen Ans=σ[d|n]d*f (d) *n/d=nσf (d)Now it's the F (n) problem. We find that for any n>=3, if X and N coprime, then n-x must be with n coprimeTherefore, the number of N and n coprime can be 22 and the number of n pairs, a total of φ (n)/2 pairs, so f (n) =n*φ (n)/2Note that the above

Number theory. Http://acm.hdu.edu.cn/showproblem.php? PID = 1, 4335 Given bpm, several N conditions are required. Proof see http://hi.baidu.com/aekdycoin/item/e493adc9a7c0870bad092fd9 The number of B, P, and m in the given question 0 With the Euler's theorem, we know that N ^ (PHI (p) has 1 (mod P), but gcd (n, p) = 1 is required here. Obviously, the question

Test instructions: Given the initial value of the For loop, the end value and increment, and a modulo, the minimum number of cycles.Analysis:After reading the question should know is a concept of congruence, so it is to solve a one-time congruence equation, like the above problem with the expansion of Euclidean algorithm. The trick point of the problem is K Max is 32, then 2^32 out of int, to use a long long, so in 1Code:#include Copyright NOTICE: Th

Original link https://www.cnblogs.com/zhouzhendong/p/CF542D.htmlQuestion portal-cf542d question portal-51nod1477 Define the formula $ J (x) = \ sum _ {1 \ Leq k \ Leq X and k | X and \ gcd (K, X/K) = 1} k $. Given an integer $ A $, the number of positive integers $ x $ must be equal to $ J (x) = A $. $ X | N $ indicates that $ x $ is a factor of $ N $. $ \ Gcd (a, B) indicates the maximum common divisor of $ A $ and $ B $. $1 \ Leq A \ Leq 10 ^ {12} $

Order Status $ f (I, j) $ modulo $ d $ is $ I $, and is the minimum number for $ J $. You can use $ BFS $ to transfer But it's gone... Complexity $ O (10ds) $ #include Codeforces1070a find a number Graph Theory

End; thee:=Next[e]; + End; - End; $ $ functionmi (x,y:int64): Int64; - varTmp:int64; - begin themi:=1; tmp:=x; - whileY>0 DoWuyi begin the ifY and 1=1 ThenMi:=mi*tmpMoDmo; -Tmp:=tmp*tmpMoDmo; WuY:=y>>1; - End; About End; $ - functionexf (X:int64): Int64; - begin -Exit (MI (x,mo-2)); A End; + the begin -Assign (input,'forest.in'); Reset (input); $Assign (output,'Forest.out'); Rewrite (output); the READLN (n); the fori:=1 toN Do the begin the read (a[i]); -f[i]:=A[i]; in End; t

: There are many solutions to the indefinite equations calculated by expanding the Euclidean, the final inverse should be x%n, that is, the range of the inverse is [0,n-1]Multiply inverse of ax≡1 (mod n), if not present, returns-1Long long inverse (long long a,long longn){Long long x, y;if (EXTENDED_GCD (a,n,x,y) ==1) return (x+n)%n; else return-1;}After EXTENDED_GCD () is executed, x may be negative, and N will become an integer. TakeModulo operations have a rate of allocation for addition, sub